How do you express $\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x}$ $(2x^2-18x-12)/(x^3-4x)$ in partial fractions?

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How do you express $\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x}$ $(2x^2-18x-12)/(x^3-4x)$ in partial fractions?

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Answer 1 · 2016-10-30T06:43:37

$\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x} = \frac{3}{x} + \frac{4}{x + 2} - \frac{5}{x - 2}$ $(2x^2 - 18x - 12)/(x^3 - 4x) = 3/x + 4/(x + 2) - 5/(x - 2)$

Explanation:

Expand:
$\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 2}$ $(2x^2 - 18x - 12)/(x^3 - 4x) = A/x + B/(x + 2) + C/(x - 2)$

Multiply both sides by $x (x + 2) (x - 2)$ $x(x + 2)(x - 2)$

$2 x^{2} - 18 x - 12 = A (x + 2) (x - 2) + B (x) (x - 2) + C (x) (x + 2)$ $2x^2 - 18x - 12 = A(x + 2)(x - 2) + B(x)(x - 2) + C(x)(x + 2)$

Let x = 0 to make B and C disappear:

$- 12 = A (2) (- 2)$ $-12 = A(2)(-2)$

$A = 3$ $A = 3$

Let x = -2 to make A and C disappear:

$2 {(- 2)}^{2} - 18 (- 2) - 12 = B (- 2) (- 2 - 2)$ $2(-2)^2 - 18(-2) - 12 = B(-2)(-2 - 2)$

$32 = B (8)$ $32 = B(8)$

$B = 4$ $B = 4$

Let x = 2 to make A and B disappear:

$2 {(2)}^{2} - 18 (2) - 12 = C (2) (2 + 2)$ $2(2)^2 - 18(2) - 12 = C(2)(2 + 2)$

$- 40 = C (8)$ $-40 = C(8)$

$C = - 5$ $C = -5$

$\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x} = \frac{3}{x} + \frac{4}{x + 2} - \frac{5}{x - 2}$ $(2x^2 - 18x - 12)/(x^3 - 4x) = 3/x + 4/(x + 2) - 5/(x - 2)$

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How do you express $\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x}$ $(2x^2-18x-12)/(x^3-4x)$ in partial fractions?

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How do you express 2x2−18x−12x3−4x(2x^2-18x-12)/(x^3-4x) in partial fractions?

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How do you express $\frac{2 x^{2} - 18 x - 12}{x^{3} - 4 x}$ $(2x^2-18x-12)/(x^3-4x)$ in partial fractions?